Let $(A_{n},*)$ denote the $n$-th classical Laver table. Let $X_{n}$ be the set of all finite sequences of elements from $A_{n}$.

Define a function $E_{n}:X_{n}\rightarrow X_{n}$ by letting

$E_{n}((x))=(x)$.

$E_{n}((2^{n},x_{1},...,x_{k}))=(x_{1},...,x_{k})$

$E_{n}((x,1,x_{1},...,x_{k}))=(x+1,x_{1},...,x_{k})$ whenever $x_{1},...,x_{k}\in A_{n}$ and $x<2^{n}$

$E_{n}((x,y,x_{1},...,x_{k}))=(x,y-1,x+1,x_{1},...,x_{k})$ whenever $x<2^{n}$ and $y>1$.

The motivation behind the function $E_{n}$ is that if $E_{n}(x_{1},...,x_{r})=(y_{1},....,y_{s})$, then $(...(x_{1}*x_{2})...)*x_{r}=(...(y_{1}*y_{2})...)*y_{s}$. Furthermore, for all $x_{1},...,x_{r}$, there is some $m$ with $E_{n}^{m}(x_{1},...,x_{r})=((...(x_{1}*x_{2})*...)*x_{r})$. Therefore, the function $E_{n}$ represents going to a next step in an algorithm that computes the classical Laver tables (and this algorithm is simply an application of the double recursive definition of the classical Laver tables). Although this algorithm has no practical uses, a modification of this algorithm can be used to calculate Laver tables.

Let $t_{n}$ be the least natural number such that $E_{n}^{t_{n}}((x,y))=(x*y)$ whenever $x,y\in A_{n}$. The motivation behind the number $t_{n}$ is that $t_{n}$ measures the number of steps it takes to compute the product $x*y$ in $A_{n}$.

We have $t_{0}=1,t_{1}=3,t_{2}=145$ and $t_{3}=599$ and $t_{4}>15000000$.

Can anyone give any good bounds on the function $n\mapsto t_{n}$? Is the function $n\mapsto t_{n}$ primitive recursive?